Unraveling the Mystery of Solving Square Roots

In summary, the conversation is discussing a mathematical equation involving square roots and trying to figure out why one method does not work. The correct method is shown and it is concluded that both solutions presented are valid.
  • #1
TSN79
424
0
I'm just wondering about something, look at this:

[tex]
\[
\begin{array}{l}
\sqrt {3 + x} + \sqrt {2 - x} = 3 \\
\left( {\sqrt {3 + x} } \right)^2 + \left( {\sqrt {2 - x} } \right)^2 = 3^2 \\
3 + x + 2 - x = 9 \\
5 = 9 \\
\end{array}
\]
[/tex]

Obviously something isn't right here, and I know that if I isolate one of the square roots on the other side, then it all works out. But why doesn't this way work?!
 
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  • #2
Why should it??
What you're basically saying in line 2, is that (a+b)^2=a^2+b^2
 
  • #3
Yes, take a look at line two, you have to square the entire left side of the equation.
 
  • #4
TSN79, remember that [tex] ( a + b ) ^ 2 = a^2 + 2ab + b^2 [/tex] :redface:

Therefore, fixing the second line, you might have:

[tex] \sqrt {3 + x} + \sqrt {2 - x} = 3 \Rightarrow 2\sqrt {\left( {3 + x} \right)\left( {2 - x} \right)} = 4 \Rightarrow [/tex]

[tex] \sqrt {\left( {3 + x} \right)\left( {2 - x} \right)} = 2 \Rightarrow \left( {3 + x} \right)\left( {2 - x} \right) = 4 \Rightarrow [/tex]

[tex] x^2 + x - 2 = 0 \Rightarrow \left( {x + 2} \right)\left( {x - 1} \right) = 0 \Rightarrow \boxed{x = - 2,1} [/tex]
 
Last edited:
  • #5
That's wrong, bomba923. This line is wrong:
bomba923 said:
[tex] 4 - x^2 = 0 \Rightarrow x = \pm 2 [/tex]
It should look like this:
[tex]x ^ 2 + x - 2 = 0 \Rightarrow \left[ \begin{array}{l} x = 1 \\ x = - 2 \end{array} \right.[/tex] :wink:
Viet Dao,
 
  • #6
VietDao29 said:
That's wrong, bomba923. This line is wrong:

It should look like this:
[tex]x ^ 2 + x - 2 = 0 \Rightarrow \left[ \begin{array}{l} x = 1 \\ x = - 2 \end{array} \right.[/tex] :wink:
Viet Dao,
Indeed! :redface:, stupid mistakes are the bane of my math education!!

Anyway,
Thanks for pointing that out-->
Mistake correct, see updated post! :smile:
-------
 
Last edited:
  • #7
--------------------
However, :blushing:
[tex] \because \forall x > 2,\;\sqrt {2 - x} \notin \mathbb{R} \, , {so} [/tex]
[tex] \therefore \boxed{x = - 2} [/tex]​

Both of your solutions are admissible.
 
  • #8
Curious3141 said:
Both of your solutions are admissible.
Oh yes, that's right! [tex] 1 < 2 [/tex], :cool:
See updated post!
 

What is a square root mystery?

A square root mystery is a problem or puzzle involving finding the number that, when multiplied by itself, equals a given number. It is called a mystery because the number is not explicitly stated and must be solved through a series of steps.

How do you solve a square root mystery?

To solve a square root mystery, you can use a variety of methods such as prime factorization, the square root algorithm, or using a calculator. The specific method will depend on the complexity of the problem and the tools available.

Why are square root mysteries important?

Square root mysteries are important because they develop critical thinking skills and problem-solving abilities. They also have practical applications in fields such as mathematics, engineering, and science.

What is the difference between a square root and a perfect square?

A square root is the inverse operation of squaring a number, while a perfect square is a number that is the result of squaring an integer. In other words, a square root finds the number that, when multiplied by itself, equals a given number, while a perfect square is a number that can be written as the product of two equal integers.

Can square root mysteries have more than one solution?

Yes, square root mysteries can have more than one solution. This is due to the fact that a number can have two square roots, a positive and a negative. However, in most cases, the positive value is considered the principal square root.

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